Let $f(x, y) = \ln(y)$. Suppose $\vec{a} = (3, 4)$ and $\vec{v} = \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right)$. Find the directional derivative of $f(x, y)$ at $\vec{a}$ in the direction of $\vec{v}$. $\dfrac{\partial f}{\partial v} = $
Explanation: The directional derivative of a function $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $\left( \nabla f (\vec{a}) \right) \cdot \vec{v}$. Let's find the gradient of $f$. $\nabla f = \left( \dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y} \right) = \left( 0, \dfrac{1}{y} \right)$ Therefore: $\begin{aligned} \left( \nabla f (\vec{a}) \right) \cdot \vec{v} &= \left( 0, \dfrac{1}{4} \right) \cdot \left( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2} \right) \\ \\ &= \dfrac{1}{8} \end{aligned}$ In conclusion, the directional derivative of $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $\dfrac{1}{8}$.